Finding freq., Wavelentgh, Phase Velocity, and attenuation constant

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Homework Help Overview

The discussion revolves around a wave equation given in the form of \( v(z, t) = 5 e^{-\alpha z} \sin(4\pi \times 10^9 t - 20\pi z) \). Participants are tasked with finding the frequency, wavelength, phase velocity, and attenuation constant based on this equation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between wave parameters such as frequency, wavelength, and phase velocity using the provided wave equation. There are attempts to derive the attenuation constant from the amplitude at a specific distance.

Discussion Status

Some participants have offered insights into the interpretation of the attenuation constant and its dependence on the wave's amplitude. Questions have been raised regarding the elimination of time from the calculations and the role of the sine term in the context of the amplitude.

Contextual Notes

Participants are navigating the implications of the wave equation's terms and their physical meanings, particularly in relation to the amplitude and its decay over distance. There is an ongoing examination of how different components of the equation interact with each other.

VinnyCee
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Homework Statement



Given...

v\left(z,\,t\right)\,=\,5\,e^{-\alpha\,z}\,sin\left(4\pi\,\times\,10^9\,t\,-\,20\pi\,z\right)

where z is distance (m), find...

(a) Frequency

(b) Wavelength

(c) Phase Velocity

(d) At z = 2m, the amplitude is 1 [V], Find the attenuation constant (\alpha).

Homework Equations



f\,=\,\frac{1}{T}

y\left(x,\,t\right)\,=\,A\,cos\left(\frac{2\pi\,t}{T}\,-\,\frac{2\pi\,x}{\lambda}\,+\,\phi_0\right)

u_p\,=\,f\,\lambda

The Attempt at a Solution

(a)

Using the first term (\frac{2\pi\,t}{T}) in the argument to the cosine in the general form above...

\frac{2\pi}{T}\,=\,4\pi\,\times\,10^9\,\,\longrightarrow\,\,T\,=\,\frac{2\pi}{4\pi\,\times\,10^9}\,=\,0.5\,\times\,10^{-9}

f\,=\,\frac{1}{T}\,=\,\frac{1}{0.5\,\times\,10^{-9}}\,=\,2\,\times\,10^9\,=\,2\,Ghz(b)

Using the second term (-\,\frac{2\pi\,x}{\lambda}) in the argument to the cosine in the general form above...

\frac{2\pi}{\lambda}\,=\,20\pi\,\,\longrightarrow\,\,\lambda\,=\,\frac{2\pi}{20\pi}\,=\,\frac{1}{10}\,=\,0.1\,m(c)

u_p\,=\,f\,\lambda\,=\,\left(2\,\times\,10^9\right)\,(0.1)\,=\,200,000,000\,\frac{m}{s}(d)

1\,=\,5\,e^{-2\,\alpha}\,sin\left(4\pi\,\times\,10^9\,t\,-\,40\pi\right)

5\,e^{-2\alpha}\,=\,1\,\,\longrightarrow\,\,-2\alpha\,=\,ln\left(\frac{1}{5}\right)\,\,\longrightarrow\,\,\alpha\,=\,0.8047

Right?
 
Last edited:
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How did you eliminate 't' in part d?
 
There is no "t" in (d). The "attenuation" constant is the rate at which the magnitude of the wave degrades- and that depends entirely upon the coefficient of the cosine term, 5e^{-\alpha z}. And here, we are given that z= 2.
 
What about the sin term. 1= 5xexp(-alpha x z) x sin term which contains t?
 

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